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In a YDSE, D = 1 M, d = 1 mm, and lambda...

In a YDSE, `D = 1 M, d = 1 mm, and lambda = 1//2 mm.`
(a) Find the distance between the first and central maxima on the screen.
(b) Find the number of maxima and minima obtained on the screen.

Text Solution

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a. `D gt gt d`
Hence , path difference at any angular position theta on the screen
`Delta x = d sin theta`
The path diference for first maxima
`Delta x = d sin theta = lambda implies sin theta = (lambda)/(d) = (1)/(2)`
`theta = 30^(@)`
Hence, distance between central maxima and first maxima
`y = D tan theta = (1)/(sqrt3) m`
b. Maximum path difference, `Delta x_(max) = d = 1 mm`
`implies` Hightest order maxima, `n_(max) = [(d)/(lambda)] = 2` and highest
order minima `n_(min) = [(d)/(lambda) + (1)/(2)] = 2`
Total number of maxima `=2n_(max) + 1 = 5`
Total number of minima `= 2n_(min) = 4`
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